algebra precalculus - time speed distance

Two horses start simultaneously towards each other and meet after $3h 20 min$. How much time will it take the slower horse to cover the whole distance if the first arrived at the place of departure of the second $5 hours$ later than the second arrived at the departure of the first.

MY TRY::

Let speed of 1st be a kmph and 2nd be b kmph

Let the distance between A and B be d km

d = 10a/3 + 10b/3

and

d/a - d/b = 5

now i cant solve it. :(

Spoiler: The answer is $10$ hours.

Answer

Let the distance from the meeting place of the departure of the first Horse $A$ is $a $ meter
and the distance from the meeting place of the departure of the first Horse $B$ is $b$ meter

So, the total distance is $a+b$ meter

So, the speed of the first horse is $\displaystyle\frac a{200}$ meter/minute and that of the second is $\displaystyle\frac b{200}$ meter/minute

So, the first horse $A$ will need to cover $b$ meter more which it will take $\displaystyle\frac b{\frac a{200}}=\frac{200b}a$ minute

So, the total time taken by $A$ will be $\displaystyle200+\frac{200b}a$ minute

Similarly, the total time taken by $B$ will be $\displaystyle200+\frac{200a}b$ minute

If $A$ is slower than $B,$ $\displaystyle200+\frac{200b}a-\left(200+\frac{200a}b\right)=300\implies 2b^2-3ab-2a^2=0\implies b=2a$ (why?)

The total time taken by $A$ will be $\displaystyle\frac{a+b}{\frac a{200}}$ minute